Evaluate $\int^{\pi/2}_0\cos^2x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac\pi4$ (Choice B) B $\dfrac\pi2$ (Choice C) C $\pi$ (Choice D) D $2\pi$
Since this has only an even power of $~\cos x\,$, we begin by using the identity $~\cos^2x=\dfrac{1+\cos(2x)}2\,$. $ \int^{\pi/2}_0\cos^2x\,dx=\dfrac12\int^{\pi/2}_0\big(1+\cos(2x)\big)\,dx$ We can now evaluate this integral. $ \begin{aligned}\int^{\pi/2}_0\cos^2x\,dx&=\dfrac12\int^{\pi/2}_0\big(1+\cos(2x)\big)\,dx\\ \\ \\&=\dfrac12\Big(x+\dfrac12\sin(2x)\Big)\Bigg|_0^{\tfrac\pi2}\\ \\ \\&=\dfrac12\Big(\dfrac\pi2+0\Big)-\dfrac12\big(0+0\big)\\ \\ \\ &=\frac\pi4\end{aligned}$